Question: Solve for $x$. Enter the solutions from least to greatest. $(x - 1)^2 - 9 = 0$ $\text{lesser }x = $
Solution: $\begin{aligned} (x - 1)^2 - 9 &= 0 \\\\ (x-1)^2&=9 \\\\ \sqrt{(x-1)^2}&=\sqrt{9} \end{aligned}$ $\begin{aligned} x-1&=\pm3 \\\\ x&=\pm3+1 \\ \phantom{(x - 1)^2 - 9}& \\ x=-2&\text{ or }x=4 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= -2 \\\\ \text{greater } x &= 4 \end{aligned}$